The use of social networks has grown dramatically all over the world. In a recent sample of 24 American social network users and each was asked for the amount of time spent (in hours) social networking each day. The mean time spent was 3.19 hours with a standard deviation of 0.2903 hours. Find a 99% confidence interval for the true mean amount of time Americans spend social networking each day

Answers

Answer 1

Answer:

Answer:

The 99% confidence interval for the true mean amount of time Americans spend social networking each day is (3.02 hours, 3.36 hours).

Step-by-step explanation:

The (1 - α)% confidence interval for population mean when the population standard deviation is not known is:

$CI=\bar x\pm t_(\alpha/2, (n-1))* (s)/(√(n))$

The information provided is:

$n=24\n\bar x=3.19\ \text{hours}\ns=0.2903\ \text{hours}$

Confidence level = 99%.

Compute the critical value of t for 99% confidence interval and (n - 1) degrees of freedom as follows:

$t_(\alpha/2, (n-1))=t_(0.01/2, (24-1))=t_(0.005, 23)=2.807$

*Use a t-table.

Compute the 99% confidence interval for the true mean amount of time Americans spend social networking each day as follows:

$CI=\bar x\pm t_(\alpha/2, (n-1))* (s)/(√(n))$

$=3.19\pm 2.807* (0.2903)/(√(24))\n\n=3.19\pm 0.1663\n\n=(3.0237, 3.3563)\n\n\approx (3.02, 3.36)$

Thus, the 99% confidence interval for the true mean amount of time Americans spend social networking each day is (3.02 hours, 3.36 hours).

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Answers

Answer:

The answer is 2

Step-by-step explanation:

Use a calculator.

Marie is reading a 271-page book. She has already read 119 pages. She uses the equation 119+8h=271 to find out how long it will take her to finish the book if she reads at about 8 pages per hour.Solve for h.

Answers

Answer: the answer is 19

Step-by-step explanation:okay so she already have read 119 pages and the total pages is 271 so we substract 271-119=152 so she has 152 pages left so we divide 152 pages divided by 8 because she read 8 pages per hour so 152/8=19

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of length 4. The number of binary code words, B(n), of length n, is shown above. If the length is increased from n to n+1, how many more binary code words will there be? The answer is 2^n, but I don't get how they got that answer. I would think 2^n+1 minus 2^n would be 2. Please help me! Thank you!

Answers

Answer:

More number of words that can be made: $\bold{2^n}$

Please refer to below proof.

Step-by-step explanation:

Given that:

The number of binary code words that can be made:

$B(n) =2^n$

where n is the length of binary numbers.

Binary numbers means 2 possibilities either 0 or 1.

Here, suppose if we have 5 as the length of binary number.

And there are 2 possibilities for each digit.

So, total number of possibilities will be $2* 2* 2* 2* 2 = 2^5$

If the length of binary number is 2.

The total words possible are $2^2$ .

These numbers are:

{00, 01, 10, 11}

If the length of binary number is 3. (increasing the 'n' by 1)

The total words possible are $2^3$ .

These words are:

{000, 001, 010, 100, 011, 101, 110, 111}

So, number of More binary words = 8 - 4 = 4 or $2^2$ or $2^n$ .

So, the answer is $2^n$ .

Let us try to prove in generic terms:

$B(n) = 2^n$

Increasing the n by 1:

$B(n+1) = 2^(n+1)$

Number of more words made by increasing n by 1:

$B(n+1) -B(n)= 2^(n+1) -2^n\n\Rightarrow 2* 2^(n) -2^n\n\Rightarrow 2^n(2-1)\n\Rightarrow \bold{2^n}$

Hence, proved that:

More number of words that can be made: $\bold{2^n}$

Final answer:

When the length of a binary code word increases from n to n+1, the number of additional binary code words is equal to the number of binary code words of length n, which is $2^n$ .

Explanation:

When the length is increased from n to n+1, the number of binary code words of length n+1 is equal to the number of binary code words of length n multiplied by 2. This is because for each binary code word of length n, we can append a 0 or a 1 to create two new binary code words of length n+1. Therefore, the number of additional binary code words is equal to the number of binary code words of length n, which is $2^n$ .

Learn more about Binary code words here:

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